Difference between revisions of "2007 AMC 10A Problems/Problem 10"
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== See also == | == See also == | ||
Revision as of 10:47, 16 February 2016
Contents
Problem
The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is 
, the father is 
years old, and the average age of the mother and children is 
. How many children are in the family?
Solution 1
Let 
be the number of children. Then the total ages of the family is 
, and the total number of people in the family is 
. So
![\[20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6\ \mathrm{(E)}.\]](http://latex.artofproblemsolving.com/1/b/e/1be138d16be316ae6222625ea479d7d88a1e76f9.png)
Solution 2
Let x be number of children+the mom. The father, who is 48, plus the number of kids and mom divided by the number of kids and mom plus 1 (for the dad)=20. This is because the average age of the entire family is 20. Basically, this looks like 48+16x/x+1=20 48+16x=20x+20 4x=28 x=7
7 children + 1 mom = 6 children.
E is the answer
See also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
